Exact differential equations

1) Solve the initial value problem (x - xy) + (y + x2) dy/dx = 0, y(0)=2
Hint: The ODE is not exact, but can be made exact. There exists an integrating factor of the form u=u(x2+y2)

M(x,y) + N(x,y) y' = 0

Omitting the hint, I found out that
(Nx - My) / M = 3 / (1-y) which is just a function of y

So u(y) = exp∫ 3 / (1-y) dy = 1 / |1-y|3 must be an integrating factor which can make the given ODE exact.

But how can I cope with the absolute value? I can't integrate until I can get rid of it. Now, if I just take one of u(y) = 1 / (1-y) or u(y) = 1 / (y-1), is that OK? Will I be getting the same general solution?

Another thing is that the initial condition is x=0, y=2>1, so must we take u(y) = 1 / (y-1) and not u(y) = 1 / (1-y)?

Secondly, I am very interested in what integrating factor the hint is referring to, what does it mean and how can I find it?

1) Solve the initial value problem (x - xy) + (y + x2) dy/dx = 0, y(0)=2
Hint: The ODE is not exact, but can be made exact. There exists an integrating factor of the form u=u(x2+y2)

M(x,y) + N(x,y) y' = 0

Omitting the hint, I found out that
(Nx - My) / M = 3 / (1-y) which is just a function of y

So u(y) = exp∫ 3 / (1-y) dy = 1 / |1-y|3 must be an integrating factor which can make the given ODE exact.

But how can I cope with the absolute value? I can't integrate until I can get rid of it. Now, if I just take one of u(y) = 1 / (1-y) or u(y) = 1 / (y-1), is that OK? Will I be getting the same general solution?

Did you try it and see?

Another thing is that the initial condition is x=0, y=2>1, so must we take u(y) = 1 / (y-1) and not u(y) = 1 / (1-y)?

Again, why not try it? Certainly one would suspect, since you would up with a "1- y" in the denominator, that solutions will not cross the line y= 1. You might check to see if that is true. If so, then, since y(0)> 1 you could assume that 1-y is always negative and |1- y|= y- 1.

Secondly, I am very interested in what integrating factor the hint is referring to, what does it mean and how can I find it?

For the third time try- don't be afraid to just plug things in and see what happens. If v(x,y) is an "integrating factor", then multplying by it makes the equation exact. If you think that the integrating factor is a function of [itex]x^2+ y^2[/itex], try multiplying by [itex]f(x^2+ y^2)[/itex] and see what happens.

Yes, I got the same answer no matter I take |1-y|=1-y or |1-y|=y-1. In general, will they always be the same? (If so, then I don't have to waste time checking both cases every time)

Again, why not try it? Certainly one would suspect, since you would up with a "1- y" in the denominator, that solutions will not cross the line y= 1. You might check to see if that is true. If so, then, since y(0)> 1 you could assume that 1-y is always negative and |1- y|= y- 1.

OK, so for this entire question, we should take |1- y|= y - 1

For the third time try- don't be afraid to just plug things in and see what happens. If v(x,y) is an "integrating factor", then multplying by it makes the equation exact. If you think that the integrating factor is a function of [itex]x^2+ y^2[/itex], try multiplying by [itex]f(x^2+ y^2)[/itex] and see what happens.

You mean Ny, right?
f is a function of two variables, x and y, so this should give rise to a "partial" differential equation, right? But when you write only f on the last line, it seems like just one variable, is this a valid step?
So in order to find the integrating factor u(x2+y2), I need to solve another ODE, right?

No, you do not get a partial differential equation. f is a function of a single variable, say f(u), and we have replaced u by x2+ y2.
The derivate, with respect to x, of f(x2+ y2), is 2x f ' (x2+ y2 where f ' is the ordinary derivative of f. I have used the chain rule: df/dx= (df/du)(du/dx).